A simply supported RCC beam of cross section $0.4 \text{ m} \times 0.6 \text{ m}$ covers a span of $8 \text{ m}$. It is subjected to a uniformly distributed load of $30 \text{ kN/m}$. If the unit weight of concrete is $24 \text{ kN/m}^3$, the tensile stress (in $N/mm^2$, rounded off to two decimal places) at the bottom of the beam at mid-span is______
RCC Beam Tensile Stress Calculation
The question asks for the tensile stress at the bottom of a simply supported RCC beam at mid-span. This involves calculating the total load, the maximum bending moment, and then the bending stress using the section modulus.
Calculate Beam Self-Weight
The self-weight of the beam contributes to the total load. It is calculated per meter length.
- Beam dimensions: width ($b$) = 0.4 m, depth ($d$) = 0.6 m
- Unit weight of concrete ($\gamma_c$) = 24 kN/m3
- Self-weight per meter ($w_{self}$):
$w_{self} = b \times d \times \gamma_c$ $w_{self} = 0.4 \text{ m} \times 0.6 \text{ m} \times 24 \text{ kN/m}^3$ $w_{self} = 5.76 \text{ kN/m}$
Calculate Total Uniformly Distributed Load (UDL)
The total load is the sum of the applied load and the beam's self-weight.
- Applied UDL ($w_{applied}$) = 30 kN/m
- Total UDL ($w_{total}$):
$w_{total} = w_{applied} + w_{self}$ $w_{total} = 30 \text{ kN/m} + 5.76 \text{ kN/m}$ $w_{total} = 35.76 \text{ kN/m}$
Calculate Maximum Bending Moment
For a simply supported beam subjected to a UDL, the maximum bending moment occurs at the mid-span.
- Span ($L$) = 8 m
- Maximum Bending Moment ($M_{max}$):
$M_{max} = \frac{w_{total} \times L^2}{8}$ $M_{max} = \frac{35.76 \text{ kN/m} \times (8 \text{ m})^2}{8}$ $M_{max} = 35.76 \text{ kN/m} \times 8 \text{ m}$ $M_{max} = 286.08 \text{ kNm}$
Calculate Section Modulus
The section modulus ($Z$) for a rectangular section is needed to calculate bending stress.
- Section Modulus ($Z$):
$Z = \frac{b \times d^2}{6}$ $Z = \frac{0.4 \text{ m} \times (0.6 \text{ m})^2}{6}$ $Z = \frac{0.4 \times 0.36}{6} \text{ m}^3$ $Z = 0.024 \text{ m}^3$
Calculate Tensile Stress at Mid-span Bottom
The tensile stress is calculated using the bending stress formula. Tensile stress occurs at the bottom fiber of the beam under positive bending moment.
- Bending Stress ($\sigma$):
$\sigma = \frac{M_{max}}{Z}$ $\sigma = \frac{286.08 \text{ kNm}}{0.024 \text{ m}^3}$ $\sigma = 11920 \text{ kN/m}^2$
- Unit Conversion to $N/mm^2$:
$1 \text{ kN/m}^2 = 1000 \text{ N} / (1000 \text{ mm})^2 = 0.001 \text{ N/mm}^2$ $\sigma = 11920 \times 0.001 \text{ N/mm}^2$ $\sigma = 11.92 \text{ N/mm}^2$
- Rounding off to two decimal places gives $11.92 \text{ N/mm}^2$.
The calculated tensile stress is $11.92 \text{ N/mm}^2$, which lies between 11.8 and 12.
Important Questions from Strength of Materials
- A simply-supported steel beam made of an I-section has a span of $8 \text{ m}$. The beam is carrying a uniformly distributed load of $15 \text{ kN/m}$. The overall depth of the beam is $450 \text{ mm}$. The moment of inertia of the beam section is $18000$ cm$^4$. The maximum bending stress in the beam will be _________ N/mm$^2$. [in integer]
A rectangular beam section of size 300 mm (width) X 500 mm (depth) is loaded with a shear force of 600 kN. The maximum shear stress on the section in N/mm² is ___________
- A steel I-beam section is subjected to a bending moment of 96 kN-m. The moment of inertia of the beam section is $24,000 \text{ cm}^4$. The bending stress at 100 mm above the neutral axis of the beam in MPa will be ________
A simply supported beam AB has a clear span of 7 meter. The bending moment diagram (BMD) of the beam due to a single concentrated load is shown in the figure below.
The magnitude of the concentrated load in kN is __________.- A load of 30 kN is applied vertically downward at the free end of a cantilever of span 5 m. If the elastic modulus of the cantilever is 30 GPa and the section has a width of 0.3 m and a depth of 0.6 m, then, the elastic deflection (in mm) is ________
